Deckspace in Spheres: Math Calculation Question

[RainOfSteel]

Ok, now that I've got my spherical volume, I'd like to know if there is a formula that can calculate the total amount of deckspace available from all the decks of a spherical vessel.

Example: A 1000m sphere. Each Deck is 4 meters high (never mind about exceptional spaces). That's 250 decks.

If a 1000m sphere has 250 decks, how much area does their combined floor-space cover?

Is there a way to figure this mathematically?


Also, is there a way of calculating the circumerence, radius, or diameter of a deck based on how far away it is from the relative "bottom" of the sphere (as the decks are stacked up).


EDIT--05/06/2005--1607--MDT

Corrected embarressing but simple math error in calculating the number of decks. Changed 83 to 250.

 

[RandyT0001]

RoS, is that sphere 1000m in diameter? I'm having a tough time figuring out where you got 83 decks from. 1000m divided by 4m (per deck) gives me 250 decks.

Whenever I do a deckplan I draw a cutaway view on a 5' (or 10') square grid with the decks stacked in the drawing. Then I use either: count the squares from the grid/ hull circumference intersection to the centerline or use the length command on the CC2 program to get the radius of each deck.

If there is a formula, it probably involves calculus, differential equations, or quantum mechanics to figure out. All of which is beyond my understanding.

 

[far-trader]

1000m circumference maybe? Would be about right for 83 decks. Diameter about 333m for circumference of about 1047m.

Hmm, or 12m decks works out just right.

I expect there are forumlae for most or all of what you want but I don't have them memorized. It'd probably be calculus and/or geometry.

Like Randy I usually just do a quick drawing and take measurements. It's close enough and easy

No need to dig out a scientific calculator and rusty math skills

 

[RainOfSteel]

Yes, that sphere is 1000m in diameter.

See: Spherical Volume: 1000 Meter Diameter.

 

[RainOfSteel]

Originally posted by Randy Tyler:
RoS, is that sphere 1000m in diameter? I'm having a tough time figuring out where you got 83 decks from. 1000m divided by 4m (per deck) gives me 250 decks.

Oops. I'd converted 4m to 12ft mentally when plugging the number into my calculator. (1000 / 12 = 83.33).

And people wonder why I ask these questions.

You are completely correct about there being 250 decks at 4m/deck in a 1000m diameter sphere.

I've gone back and corrected my original post.

 

[RainOfSteel]

Originally posted by far-trader:

Like Randy I usually just do a quick drawing and take measurements. It's close enough and easy

No need to dig out a scientific calculator and rusty math skills

It's tough to do an off-the-cuff drawing of 250 decks.

 

[far-trader]

Originally posted by RainOfSteel:
</font><blockquote>quote:</font><hr />Originally posted by far-trader:

Like Randy I usually just do a quick drawing and take measurements. It's close enough and easy

No need to dig out a scientific calculator and rusty math skills

It's tough to do an off-the-cuff drawing of 250 decks. </font>[/QUOTE]Still easier than actually doing the deckplans for 250 decks

But yeah I see what you're saying. I'd probably get lazy and say much of the outer shell is fuel in non-cubist volume and the decks would all be cubist volume with several adjacent decks being the same footprint. So draw your circle, break it up into 25 sections of 10 equal footprint decks each (taking the smallest area for the footprint). That's how I'd tackle it. That and repetitive elements through the decks will cut down the drawing time some.

 

[vegascat]

You would not need to do 250 decks.
Engineering spaces would be grouped as decks with waste filled with fuel tankage. The decks could be near identical.
Crew areas could easily be designed as cubic spaces with identical facilities on multiple decks.
If you plan carefully, you may be able to design the whole 37MTons with as few as 25 deck drawings.

 

[far-trader]

Define "real fuel tanks"

You mean stuff like high capacity pumps, super insulation, baffles, compartmentalization, heaters, chillers, high pressure rated bulkheads, um, what else

It's even more fun when you start adding collapsible bladder fuel tanks. Or player's who want to flood a cargo deck with liquid H fuel for an extra jump

But I'm happy in my little TU where I figure the actual amount of "fuel" is half the volume and the other half of the "fuel" is the assorted equipment to actually handle it safely and make it work as noted above.

 

[RainOfSteel]

There isn't really going to be much in the way of fuel tankage.

No Jump Drives will be mounted (it's not going anywhere).

The largest liquid mass will be a cylinder of water running up from the bottom of the sphere for 500 meters, topped by a cone of open air for another 500 meters and a giant skylight. I'd guess the cylinder is 100 to 200 meters in diameter, although I'd have to make a measurement of the drawing to be sure.


EDIT--Shortly After Posting

I don't think it will have maneuver drives, either. Just contra-grav.

 

[far-trader]

Hmm, that's right, I'd forgotten this was your space arcology project. Makes things a bit harder.

 

[GypsyComet]

A sphere that size is nearly 39 million displacement tons (38.786 MDt in round numbers).

The sphere runs to about 523.6 million cubic meters. Dividing this by your chosen deck-to-deck separation of 4m gives a deck area of 130.9 million square meters.

That's a LOT of graph paper...

 

[JAFARR]

Originally posted by RainOfSteel:
</font><blockquote>quote:</font><hr />Originally posted by Randy Tyler:
RoS, is that sphere 1000m in diameter? I'm having a tough time figuring out where you got 83 decks from. 1000m divided by 4m (per deck) gives me 250 decks.

Oops. I'd converted 4m to 12ft mentally when plugging the number into my calculator. (1000 / 12 = 83.33).

And people wonder why I ask these questions.

You are completely correct about there being 250 decks at 4m/deck in a 1000m diameter sphere.

I've gone back and corrected my original post. </font>[/QUOTE]If you figure the "footprint" of a dton as 2m to a side then the deck hight is aprox. 3M. 2m x 2m x 3.5m =14 cubic m which is a dton in CT using a hight of 3.37 m = 13.48 or 13.5 cubic meters which is a dton in MT. Start in the center gives you a radus of 500m. 500/3.5 = 142.85 decks to each hemisphere. Drop the 2.85 and figure it is used as fuel tankage. That leaves a round number of 280 decks. The question then becomes at what point are the decks too small to have a useful amount of floor space.
I am half asleep, but would not the radius of each deck be the (radius of the center deck (minus the hight of a deck times the number of decks out from center?)) or rd =500-(n * dh) where rd = radius of a deck, n = deck #(Main or center deck = deck 0. Deck 1 is deck below center, Deck 01 is 1st deck above center, etc. Dh is deck hight or 3.5. (3.37 in MT) ?

 

[Uncle Bob]

You are half asleep. This is not a traveller standard spaceship with 3 or 3.5 meter (is that Gurps? CT uses 3), but a space station with 4m decks.
I too get 5.24E8 m3, for 131 million square meters, Actually a little more as the curvature means that a small fractiion on the outside will not have 4m free overhead, but on a ship this large that will be less than 1%

 

[Uncle Bob]

You are half asleep. This is not a traveller standard spaceship with 3 or 3.5 meter (is that Gurps? CT uses 3), but a space station with 4m decks.
I too get 5.24E8 m3, for 131 million square meters, Actually a little more as the curvature means that a small fractiion on the outside will not have 4m free overhead, but on a ship this large that will be less than 1%

 

[Fritz_Brown]

Easy answer (

), RoS:
In Excel (or some such), set your zoom really low to start with. Highlight the whole sheet, and make your rows and columns the exact same height/width in pixels - this gives you a square grid. Use the drawing objects to make a circle 250 rows high.

Make the thing transparent, so you can get to the boxes underneath. Put a value in the first and last boxes that are actually inside the circle in each row. Then (and you will have to zoom back in for the answers) use formulae to calculate the number of columns (remember to make it inclusive of the ones with values) for each row.

Multiply the answer by 4 (because you made your boxes=4m). Then use pie are square. Now you have the area for each deck, a simple sum will give the total. The advantage is you only write the formula once, then use auto-fill to populate the rest of the decks.

I am currently designing a belter for my Quaddies (0-g folks), that is a sphere, with 6 cylinders coming out from it. I needed to figure how much of the sphere was needed to hold some of the fuel, and used this idea to see when (going down from one "end") I had enough room. I used .1m high sections in a 7.5m radius sphere.

If you want to PM me RoS, I can pull that worksheet and e-mail it to you (may save you some work).

 

[RainOfSteel]

Originally posted by GypsyComet:
A sphere that size is nearly 39 million displacement tons (38.786 MDt in round numbers).

Yes. See: Spherical Volume: 1000 Meter Diameter.

Originally posted by GypsyComet:
The sphere runs to about 523.6 million cubic meters. Dividing this by your chosen deck-to-deck separation of 4m gives a deck area of 130.9 million square meters.

That's a LOT of graph paper...

I am making a valiant attempt to try and get Canvas 7 to help me out.

Ron Vutpadki has designed a Canvas starter page with basic elements ready-to-go for use in drawing deckplans.

But what it doesn't say is how to expand the basic page w/grid into a multi-page format that I can draw a circle 666 squares in diameter on.

Also, I'm having my doubts about whether my little podunk PC can handle Canvas 7 when there are 250 "decks", each one with a lot of pages.

However, this is really just something I'm thinking about. The size of the project is so huge, it's more than little duanting. I've got some ideas about space-usage that should save some time. Not all decks will be 4m, some will be 8m or 12m. Some compartments will be as high as 200m (and one compartment will be 500m in height, or 1km if you count the open-top water compartment directly below it).

 

[RainOfSteel]

Originally posted by Fritz88:
Easy answer (

), RoS:
In Excel (or some such), set your zoom really low to start with. Highlight the whole sheet, and make your rows and columns the exact same height/width in pixels

How do you set the height of rows and columns by pixel?

I only know Menu>Format>Row/Column, and to adjust the number in the dialog box.

When I set these both to be the same value, the grid is not square.

(Note: I have Excel 2000.)

 

[JAFARR]

Originally posted by Uncle Bob:
You are half asleep. This is not a traveller standard spaceship with 3 or 3.5 meter (is that Gurps? CT uses 3), but a space station with 4m decks.
...

I too was thinking 3m decks were standard, but then you would have to use deck squares 2.16m to a side unless the idea was 3m x 2m x 2m = 12 cubic meters and the remainder was actually consumed by the decks themselves. I have been looking for that specification, but can't seem to locate where it is given in canonized material.

Quick addition: most of the trucks (18 wheeler closed box types) you see on the road say 110 inches inside hight. That works out to 2.79m barely allowing you to put the whole box inside a 3m hight between decks. Assuming a standard cargo handling system like these land/sea/air freight companies use where they load the box once then put the box on or in what ever tyre of carrier is needed at the moment for it's next stage of the trip.

 

[RainOfSteel]

BTW . . .

The assumption operative in this is 4m from floorplate to floorplate. About 3m total ceiling clearance (like a slightly larger than normal house), and 1m for spacing between decks.

Decks at 8m and 12m would have much larger than normal ceilings. Especially like big control spaces (which wouldn't really need to be big like that, but would be built that way for stylistic reason (and the author's anime-bridge prejudices)).